題目：
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the PRocess is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
思路：
除和余
package others;public class AddDigits { public int addDigits(int num) { if (num / 10 == 0) return num; int res = 0; while (num > 0) { res += num % 10; num = num / 10; } return addDigits(res); } public static void main(String[] args) { // TODO Auto-generated method stub }}
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